I have researched the question $\lim_{n \to \infty} n*\sin(\frac{1}{n})$ quite profusely, and I know that it equals to 1, and I know why:

A) You can use a change of variables and substitute, say, $m = \frac{1}{n}$ so that $m \to 0$ instead.

B) L'Hopital's rule

The problem is, we haven't used either of these methods in class, so I am wondering if there is any other possible way to approach this question?

  • $\begingroup$ There's always Taylor expansion $\endgroup$ Commented Apr 6, 2016 at 22:42
  • $\begingroup$ Haven't really learnt that either :P $\endgroup$
    – Inazuma
    Commented Apr 6, 2016 at 22:43
  • $\begingroup$ Have you used mean value theorem? $\endgroup$
    – rtybase
    Commented Apr 6, 2016 at 22:54
  • $\begingroup$ @rtybase Pretty much all we've done is studied sequences and series (such as the ratio test, squeeze theorem etc.). The original question was actually to show whether the sum of $n*\sin(\frac{1}{n})$ diverges (or converges), and my approach was to show that because lim $n*sin(\frac{1}{n})$ doesn't equal 0, it must diverge. $\endgroup$
    – Inazuma
    Commented Apr 6, 2016 at 22:59
  • $\begingroup$ See some ideas here math.stackexchange.com/questions/75130/… $\endgroup$
    – rtybase
    Commented Apr 7, 2016 at 18:08

1 Answer 1


To prove $\sin x/x,\,\tan x/x\to 1$ as $x\to 0$, consider the areas of a small-angle sector of a circle and the right-angled triangles obtained by using a radius for a hypotenuse or base. The squeeze theorem will complete the proof since $\cos x \to 1$.


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